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Consider only the species (at standard conditions) $$\mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd}$$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?

Short Answer

Expert verified
a. The strongest oxidizing agent is \(Br_2\) (E° = 1.087 V). b. The strongest reducing agent is Ca (E° = -2.870 V). c. Br⁻, H⁺, La³⁺, Ca, and Cd can be oxidized by MnO₄⁻ in acid. d. Br₂ and H⁺ can be reduced by Zn(s).

Step by step solution

01

Find the reduction potentials of the given species

Using a standard reduction potential table, we can find the values for the given species at standard conditions: Br⁻ + e⁻ → Brˉ² | E° = -1.087 V Br₂ + 2e⁻ → 2Br⁻ | E° = 1.087 V H⁺ + e⁻ → Hˉ | E° = 0.000 V 2H⁺ + 2e⁻ → H₂(gas) | E° = 0.000 V La³⁺ + 3e⁻ → La(s) | E° = -2.380 V Ca²⁺ + 2e⁻ → Ca(s) | E° = -2.870 V Cd²⁺ + 2e⁻ → Cd(s) | E° = -0.403 V
02

Answer question a (strongest oxidizing agent)

As mentioned in the analysis, the strongest oxidizing agent is the one with the highest reduction potential. Comparing the given species' reduction potentials, we find that Br₂ has the highest reduction potential of 1.087 V. Therefore, the strongest oxidizing agent is Br₂.
03

Answer question b (strongest reducing agent)

Similarly, the strongest reducing agent is the species with the lowest reduction potential. Comparing the given species' reduction potentials, we find that Ca has the lowest reduction potential of -2.870 V. Therefore, the strongest reducing agent is Ca.
04

Find the reduction potential of MnO₄⁻ in acid

To answer question c, we need the reduction potential of MnO₄⁻ in acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | E° = 1.507 V
05

Answer question c (species oxidized by MnO₄⁻ in acid)

We can see that the reduction potential of MnO₄⁻ in acid is 1.507 V. A species can be oxidized by MnO₄⁻ if its reduction potential is lower than this value. Comparing with the given species' reduction potentials, we find that Br⁻, H⁺, La³⁺, Ca, and Cd can be oxidized by MnO₄⁻ in acid.
06

Find the reduction potential of Zn(s)

To answer question d, we need the reduction potential of Zn: Zn²⁺ + 2e⁻ → Zn(s) | E° = -0.763 V
07

Answer question d (species reduced by Zn(s))

We can see that the reduction potential of Zn(s) is -0.763 V. A species can be reduced by Zn(s) if its reduction potential is higher than this value. Comparing with the given species' reduction potentials, we find that Br₂ and H⁺ can be reduced by Zn(s).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidizing Agents
In the world of redox reactions, an oxidizing agent is a chemical species that accepts electrons from other species, causing itself to be reduced. Oxidizing agents are crucial because they facilitate the oxidation process by taking in electrons. The key to identifying a strong oxidizing agent lies in understanding reduction potentials.

Reduction potential is a measure of how readily a species gains electrons. The higher the reduction potential, the stronger the oxidizing power of a substance. This is why in our example, bromine (\(\text{Br}_2\)) is the strongest oxidizing agent among the listed species, with a reduction potential of 1.087 V. It indicates that \(\text{Br}_2\) has a high ability to accept electrons and oxidize other substances, such as reducing agents, in a chemical reaction.
Reducing Agents
Reducing agents hold an essential role in redox reactions. These species donate electrons to other substances, enabling those substances to be reduced, while the reducing agent itself becomes oxidized. Essentially, reducing agents drive the reduction process forward.

To determine the strength of reducing agents, we look at their reduction potentials. A strong reducing agent will have a very low or even negative reduction potential. In the given example, calcium (\(\text{Ca}\)) stands as the strongest reducing agent with a reduction potential of -2.870 V. This negative value means that \(\text{Ca}\) readily donates electrons to other species, enhancing their reduction. So, whenever interacting with other substances with higher reduction potentials, \(\text{Ca}\) will be eager to give up its electrons.
Reduction Potential
Understanding reduction potential is crucial in any discussion involving redox reactions. Reduction potential, symbolized as \(E^° \), indicates how likely a chemical species is to gain electrons and be reduced. The higher the \(E^° \) value, the greater the tendency to accept electrons.

Reduction potentials are usually presented in volts and are determined under standard conditions. By comparing these values, one can predict the direction of electron flow in redox reactions. For example, \(\text{MnO}_4^-\) has a reduction potential of 1.507 V in acidic conditions, making it a powerful oxidizing agent capable of oxidizing species with lower reduction potentials. This concept is pivotal in predicting and balancing redox reactions efficiently.
Standard Conditions
Redox reactions' measurements are often taken under standard conditions. These conditions, typically identified as 25°C (298 K), 1 atm pressure, and 1 M concentration for solutions, ensure that all reactions and potentials are measured consistently.

Under these conditions, the standard electrode potential for each species can be compared reliably. This consistency allows chemists to predict the feasibility of reactions and understand the behavior of electron transfer between species in redox reactions. Having a standard reference is vital as it permits benchmarking of reduction and oxidation capabilities of various agents, such as analyzing those covered: \(\text{Br}_2\), \(\text{H}^+\), and \(\text{Ca}\).

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Most popular questions from this chapter

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{~kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

Consider the following half-reactions: $$\begin{array}{cc}\mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 \mathrm{M}\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\) c. \(\begin{aligned} \mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Acid}}{\longrightarrow} \\ & \mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q) \\ \text { d. } \mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow} \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { e. } \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow} \\\ \mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \end{aligned}\)

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